Question

An object having velocity 4 m /s is accelerated at the rate of 1.2m/s^2for 5 sec. Find the distance travelled during the period of acceleration

Answer 1

HOLA! !!

HERE WE GO

GIVEN

initial velocity (u) = 4ms^-1

acceleration (a) = 1.2ms^-2

time (t) = 5s


SOLUTION

we know a kinematical equation

s = ut +1/2at ^2

= 4 (5) +1/2 (1.2)(5)^2

= 20 + (0.6)(25)

= 20 + 15

= 35m

there distance = 35m

HOPE IT HELPS YOU!!!

Answer 2

Answer:

Refer to the attachment

Distance is 35m.